Question

Golf course designer Roberto Langabeer is evaluating two sites, Palmetto Dunes and Ocean Greens, for his next golf course. He wants to prove that Palmetto Dunes residents (population 1) play golf more often than Ocean Greens residents (population 2). Roberto plans to test this hypothesis using a random sample of 81 individuals from each suburb. His alternative hypothesis is __________.

Answer 1

Answer:

a) Null hypothesis:$\mu_{1} \leq \mu_{2}$

Alternative hypothesis:$\mu_{1} > \mu_{2}$

b) $z_{crit}=2.33$

And since the calculated value is lower than the critical value we have enough evidence at 0.01 of significance to FAIL to reject the null hypothesis.

Step-by-step explanation:

Part a

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

We want to test this:

Null hypothesis:$\mu_{1} \leq \mu_{2}$

Alternative hypothesis:$\mu_{1} > \mu_{2}$

Part b

Golf course designer Roberto Langabeer is evaluating two sites, Palmetto Dunes and Ocean Greens, for his next golf course. He wants to prove that Palmetto Dunes residents (population 1) play golf more often than Ocean Greens residents (population 2). Roberto commissions a market survey to test this hypothesis. The market researcher used a random sample of 64 individuals from each suburb, and reported the following:  X 1 = 15  times per month and  X 2 = 14  times per month. Assume that  σ 1 = 2  and  σ 2 = 3 . With  α = .01 , the critical z value is _____.

Data given and notation

$\bar X_{1}=15$ represent the mean for the sample 1

$\bar X_{2}=14$ represent the mean for the sample 2

$\sigma_{1}=2$ represent the population deviation for 1

$\sigma_{2}=3$ represent the population deviation for 2

$n_{1}=64$ sample size selected for 1

$n_{2}=64$ sample size selected for 2

$\alpha=0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

The statistic is given by:

$z=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$     (1)

$z=\frac{15-14}{\sqrt{\frac{2^2}{64}+\frac{3^2}{64}}}}=2.22$  

In order to find the critical value we need a value that accumulates 0.01 of the area on the right tail, since we are conducting a right tailed test. And the critical value is:

$z_{crit}=2.33$

And since the calculated value is lower than the critical value we have enough evidence at 0.01 of significance to FAIL to reject the null hypothesis.